Math 6802 II

Nathan Broaddus

Ohio State University

January 30, 2015 The Hom(−, B)

Definition 1 (Opposite ) If C is a category then the C op is the category with objects Ob(C ) = Ob(C op)

and for C1, C2 ∈ C morphisms

MorC op (C1, C2) = MorC (C2, C1).

The composition rule for morphisms being m2 ◦C op m1 = m1 ◦C m2.

Definition 2 A contravariant functor F : C → D is a (covariant) functor F : C → D op. Example 3 (− ⊗ B is covariant) Fix the B. The functor F : Ab → Ab with F (A) = A ⊗ B and F (h) = h ⊗ 1B is covariant.

Example 4 (Hom(−, B) is contravariant) Fix the abelian group B. The functor G : Ab → Ab with G(A) = Hom(A, B) and G(h)(f ) = f ◦ h is contravariant.

Example 5 (Hom(A, −) is covariant) Fix the abelian group A. The functor G : Ab → Ab with G(B) = Hom(A, B) and G(h)(f ) = h ◦ f is covariant.

I We now can now repeat Lectures 3-5 with Hom(−, B) instead of − ⊗ B. ∗ I When B is fixed we will call A = Hom(A, B) the dual of A ∗ ∗ ∗ I and given f : C → D we get its dual f : D → C where f ∗(h) = h ◦ f . Proposition 6 (Properties of Hom)

1. ∼ Y Hom(⊕αAα, B) = Hom(Aα, B) α 2. ∼ M Hom(A, ⊕βBβ) = Hom(A, Bβ) β 3. Hom(Z, B) =∼ B Hom and exactness Proposition 7 (Left exactness for Hom(−, B)) Fix the abelian group B. Given exact

β C −→α D −→ E → 0

then after dualizing with B we get exact

∗ β∗ C ∗ ←−−α D∗ ←− E ∗ ← 0.

Thus Hom(−, B) is left exact (this is the definition for contravariant )

β◦− Hom(A, C) −−−→α◦− Hom(A, D) −−−→ Hom(A, E) → 0

is exact (thus Hom(A, −) is right exact).

1. Prove the left and right exactness statements above. 2. Show that the opposite exactness statements do not hold in general. Hom(−, B) and chain complexes

I Recall that a (C∗, ∂) is a graded abelian group M C∗ = Cn n∈Z

together with a degree -1 homomorrphism ∂ : C∗ → C∗ satisfying ∂2 = 0.

Definition 8 (Cochain complex) Fix abelian group B. The cochain complex (C ∗, δ) corresponding to the chain complex (C∗, ∂) is the abelian group

∗ C = Hom(C∗, B)

together with the coboundary map δ = ∂∗ I Note that ! ∗ M Y Y ∗ C = Hom(C∗, B) = Hom Cn, B = Hom(Cn, B) = Cn n n∈Z n∈Z but we will generally only use the subgroup

M ∗ Y ∗ Cn ⊂ Cn . n∈Z n∈Z

2 2 I Also note that δ is degree 1 but we do have δ = − ◦ ∂ = 0. I In order to preserve the convention that the domain of fn is Cn we set

∗ δ = δ| ∗ = ∂ n Cn n+1

Definition 9 ( groups) Given a cochain complex (C ∗, δ) its nth cohomology group is

ker δ Hn(C ∗) = n . Im δn−1 Lemma 10 If f : C → D is a chain map then f ∗ : D∗ → C ∗ is a cochain map.

Proof. f : C → D is a chain map so ∂D f = f ∂C

∗ ∗ f ◦ δD = f (− ◦ ∂D )

= (− ◦ ∂D ) ◦ f

= − ◦ f ◦ ∂C

= δC ◦ (− ◦ f ) ∗ = δC ◦ f Lemma 11 If T : C → D is a chain homotopy between chain maps f , g : C → D then

T ∗ : D∗ → C ∗ is a cochain homotopy between chain maps f ∗ and g ∗. Note: cochain homotopies have degree -1.

Proof. T : C → D is a chain homotopy so ∂T + T ∂ = g − f

δT ∗ + T ∗δ = δ ◦ (− ◦ T ) + T ∗ ◦ (− ◦ ∂) = (− ◦ T ) ◦ ∂ + (− ◦ ∂) ◦ T = − ◦ (T ∂ + ∂T ) = − ◦ (g − f ) = (g − f )∗ = g ∗ − f ∗ Cohomology with coefficients in G

Definition 12 (Cohomology with coefficients in G) Fix the abelian group G for dualizing. Given a chain complex C let

Hn(C; G) = Hn(C ∗)

If X is a space let Hn(X ; G) = Hn(C ∗(X ))

n ∗ H˜ (X ; G) = Hn(C˜ (X ))

n n CW∗ HCW(X ; G) = H (C (X )) For (X , A) a topological pair let

Hn(X , A; G) = Hn(C ∗(X , A)) Example 13 (Cohomology of RP4) Standard CW chain complex for RP4 is

· · · → 0 → Z −−→×2 Z −−→×0 Z −−→×2 Z −−→×0 Z → 0

Dualizing (with coefficient group G = Z) we get C CW∗(RP4)

· · · ← 0 ← Z ←−−×2 Z ←−−×0 Z ←−−×2 Z ←−−×0 Z ← 0

So   Z, n = 0 n 4 ∼ HCW(RP ; Z) = Z2, n ∈ {2, 4}  0, otherwise Homology with coefficients in Z are also universal for cohomology with coefficients in G for all abelian groups G. Theorem 14 (Universal coefficient theorem for cohomology) For C a chain complex there is a split short

n 0 → Ext(Hn−1(C), G) → H (C; G) → Hom(Hn(C), G) → 0.

This sequence is natural in that if C → D is a chain map then we get commutative

n 0 / Ext(Hn−1(D), G) / H (D; G) / Hom(Hn(D), G) / 0

     n 0 / Ext(Hn−1(C), G) / H (C; G) / Hom(Hn(C), G) / 0 The Ext functor

Definition 15 (The Ext functor) Let A and B be abelian groups and F be a free of A. Dualizing F with B we get the cochain complex F ∗

Extn(A, B) = Hn(F ∗)

We showed that free resolutions of the same group are chain homotopic and the dual of a chain homotopy is a cochain homotopy. Thus we have: Theorem 16 (Ext is well-defined) Extn(A, B) is independent of the free resolution F of A. I Every abelian group A has a free resolution

∂2 ∂1 ε · · · → 0 −→ F1 −→ F0 −→ A → 0

I So we have the chain complex F

∂1 · · · → 0 → F1 −→ F0 → 0

∗ I and cochain complex F from dualizing with B

∗ δ0 ∗ · · · → 0 ← F1 ←− F0 ← 0

I  ker δ , n = 0  0 n n ∗  Ext (A, B) = H (F ) = coker δ0, n = 1   0, n 6= 0, 1

I recall given h : M → N we set coker h = N/h(M) I We have the exact sequence

∂1 ε F1 −→ F0 −→ A → 0

I Which dualizes to exact

∗ δ0 ∗ ∗ F1 ←− F0 ← A ← 0

I Which we may extend to exact

∗ δ0 ∗ ∗ 0 ← coker δ0 ← F1 ←− F0 ← A ← 0

I Hence 0 ∼ ∼ ∗ Ext (A, B) = ker δ0 = A = Hom(A, B) 1 I and Ext (A, B) fits in the exact sequence

1 ∗ δ0 ∗ 0 ← Ext (A, B) ← F1 ←− F0 ← Hom(A, B) ← 0

Definition 17

Ext(A, B) = Ext1(A, B) Example 18

Let’s compute Ext(Z60, Z42)

I Free resolution F of Z60

· · · → 0 → Z −−→×60 Z → 0

∗ I Dualizing by Z42 we get F

×60 0 ← Hom(Z, Z42) ←−− Hom(Z, Z42) ← 0

I Simplifying ×60 0 ← Z42 ←−− Z42 ← 0

I Hence

∼ ∼ Z42 Z Ext(Z60, Z42) = coker(×60) = = 60Z42 gcd(42, 60)Z Proposition 19 (Properties of Ext)

∼ Q 1. Ext(⊕αAα, B) = α Ext(Aα, B) ∼ L 2. Ext(A, ⊕βBβ) = β Ext(A, Bβ) 3. Ext(A, B) = 0 if A is free. ∼ 4. Ext(Zn, B) = B/nB Proof of Proposition 19. ∼ Q 1. Claim: Ext(⊕αAα, B) = α Ext(Aα, B) I Let Fα be a free resolution of Aα I Then ⊕αFα is a free resolution of ⊕αAα

I

1 ∗ Ext(⊕αAα, B) =∼ H ((⊕αFα) ) ∼ 1 ∗ = H (ΠαFα) ∼ Y 1 ∗ = H (Fα) α Y =∼ Ext(Aα, B) α Proof of Proposition 19. ∼ L 2. Claim: Ext(A, ⊕βBβ) = β Ext(A, Bβ)

I Let F be a free resolution of A

I

1 Ext(A, ⊕β Bβ ) =∼ H (Hom(F , ⊕β Bβ )) 1 =∼ H (⊕β Hom(F , Bβ )) M 1 =∼ H (Hom(F , Bβ )) β M =∼ Ext(A, Bβ ) β Proof of Proposition 19 (continued).

3. Claim: Ext(A, B) = 0 if A is free

I Suppose A is free.

I Use the free resolution of A

· · · → 0 → A → 0

I Ext(A, B) = coker(A → 0) = 0. ∼ 4. Claim: Ext(Zn, B) = B/nB

I Use the free resolution of Zn

· · · → 0 → Z −−→×n Z → 0 ∼  ×n  I Get Ext(A, B) = coker B −−→ B = B/nB. Universal coefficient theorem for cohomology

Theorem 20 (Universal coefficient theorem for cohomology) For C a chain complex there is a split short exact sequence

n 0 → Ext(Hn−1(C), G) → H (C; G) → Hom(Hn(C), G) → 0.

This sequence is natural in that if C → D is a chain map then we get commutative

n 0 / Ext(Hn−1(D), G) / H (D; G) / Hom(Hn(D), G) / 0

     n 0 / Ext(Hn−1(C), G) / H (C; G) / Hom(Hn(C), G) / 0